A roller-coaster car is moving at 4.12 m/s along a straight horizontal track before entering the first drop. What will the speed of the coaster be at point B? Ignore friction and air resistance. 4.00 m 25.0 m 12.0 m C 3.00 m

A 800 kg roller coaster goes over the first hill with a speed of 17.0 m/s.  If h = 58 m, determine the change in gravitational potential between A and B.  Use g = 10 N/kg.

Computation. A block is released from rest at the top of a 47-m long frictionless ramp inclined at 34° above the horizontal. What is the block's speed at the bottom of the ramp? v= m/s Record your numerical answer below, assuming three significant figures. Remember to include a "-" if/when necessary.

See photo x = 36.0 m, y = 38.2 m, z = 43.2 m a. Assuming the track is frictionless, what is the minimum speed required at Point A for the roller-coaster to reach Point B? b. If the roller-coaster starts at Point A with this speed, how fast will it be going when it reaches Point B? c. What will be the maximum speed of the roller-coaster during this part of its trip?

A skier weighing 0.85 kN goes over a frictionless circular hill as shown. If the skier's speed at point A is 8.8 m/s, what is his speed at the top of the hill (point B)?

A 4.00kg block moves with an initial speed of3.50 km/h downhillover an inclined plane as shown in the figure. The height of the inclined plane is 51.3cm and the friction coefficient between the block and the incline is 0.120. a.How much is the speed of the block at the midpoint of the inclined side. b.How much is the speed of the block at the bottom of the inclined side.

1. A roller coaster, shown below (h1 m/s = 40.0 m, h₂ = 16.0 m, h3 m), is pulled up to point A where it and its screaming occupants are released from rest. Assuming no friction, calculate the speed at point D. cci16 ceil6 ccil6 ceil6 ccil6 ce clo cci16 cci16 33 16 cci16 ccil6 ccN 6 ceil6 ccilo = 26.0 Cecil6 ceil6 ccil6 ccio cil6 cci16. 6 ccil6 cci16 cci16 ceil6 cc²

A penguin slides down an icy incline on his stomach, as shown in the diagram below. The coefficient of kinetic friction between the incline and the penguin's stomach is 0.15. The angle of the incline is 0 = 25 degrees. The total distance down the incline is 25 meters. The penguin gets a running start so his speed is 2.5 m/s when he begins to slide down the incline. What is the penguin's speed when he reaches the bottom of the incline? Hint: use Newton's 2nd law to calculate the acceleration of the penguin on the incline, then use kinematic equations to calculate the speed of the penguin when he reaches the bottom. 2.5 m/s 25 m 25°

Please see attached image for question. Thank you

m = 325 kg h1 = 19.5 m h2 = 4.7 m  A roller coaster car of mass m is rolling down the track, starting from rest, as shown in the figure. The initial height from the ground is h1 (Position-1), while the final height is h2 (Position-3). The system has no friction. What is the speed of the car when it reaches its final height (Position-3).